[SOUND] >> Welcome back to linear circuits. This is Dr. Ferri. This lesson is on transfer functions. We've looked at a circuit like this before and we were able to analyze this circuit. In terms of, what is this output for this specific input? Now, what we'd like to do in this particular lesson is make it more general. To say, all right, suppose I've got a more general form of the input. Well, I haven't specified the frequency. Or the amplitude, or even the phase. I don't want to have to go back and resolve this problem everytime I change the frequency. Or everytime I change the amplitude or phase. So, what I'd like to be able to do is come up with a more generic way. In other words, I introduce C transfer functions to find an input-to-output relationship. For the circuit that holds for all input frequencies. This is going to build upon phasors and the impedance method. Remember, what our impedences are for various components. The transfer function approach, is a way of looking at linear circuits, as if they were systems. Here ,I'm showing the issiloscope trace, with an input and an output of my circuit. So. the system point of view, is to look at the circuit like a system with an in imput and an output. Here, our input is a sinusoid, and our output is also a sinusoid, and there's a change angle. The change in angle, between the input and the output and a change in amplitude. At the same frequency. So these are the same, for the input and the output. Now, notice I show it as an x on the input. Because I want this to be kind of generic. So, my input might be able to source or current source. And my output may be a voltage or a current across a particular component. So, the transfer function is a way of characterizing that circuit and it's an H of omega. It says what do I do to my input, to give me this output. So, I want it like a function that I can use. So, I take my input and represent it in phasor form and output in the phasor form. Snd the transfer function, is a ratio of the output of the input. So, I want you to remember that, output over input. And that's my transfer function. And, it is invariant with respect to the inputs and outputs. So this particular formula, holds no matter what my inputs and outputs are. So, I could change the frequencies and this ratio will always stay following this particular formula. So, this is kind of nice because, if I know what my input output amplitude is. And I know what my transfer function is, I can find my output amplitude. And it's the same with the angles. Now let me go back, I want to go through a little bit more. Getting a better formula for you to be able to use, in actually working with problems. So, this is just showing exactly what I showed before, but this time let's look at the ratio of the amplitudes. If I'm trying to calculate what this ratio is, it's the ratio of the amplitudes. And I subtract off the phases. So, the magnitude of H, H is a complex number. The magnitude of it, is simply the ratio of the amplitudes, the output over the input amplitudes. And the angle of H, is the difference between the two phases, the output minus the input phase. Another way of writing this, again, going back to the fact that we want a formula. My output Amplitude. My output amplitude is equal to my input amplitude, times the magnitude of the transfer function. My output phase, right there, is equal to the input phase, plus the phase of the transfer function. So, this right here, very important formula, because this allows us, given a sinusoidal input. To figure out what the sinusoidal output is, if I have that transfer function as a characteristic way of representing the circuit. Let me look at an example of an RC circuit. I want to find the transfer function. Ultimately, what I'd like to be able to do is solve for V out, but the step along the way is to find that transfer function. Okay, the transfer function follows exactly on that impedance method that we’ve already done. So, I have to go through the same process that I did before, in solving circuit problems. I replace this with its impedance equivalent. So, that's V sub S, has got to be AN at an angle of theta and that's its phaser. I replace R with its impedance, which is R. And, I replace the capacitor with Z sub C. And Z sub C, is equal to one over J omega C. Okay, I want to solve for V out, I use standard methods. In this case, easiest thing to do is voltage divider. So V out is equal to Z sub C over R plus Z sub c times the input, which is AN, at an angle of theta in. And in this particular case, I do not know what omega is, so I'm going to leave that as a variable in here. And I have 1 over j omega C. R+1 over J omega C, times the input. And the output is actually going to be A out, theta out. And remember, the transfer function is the ratio of the output over the input. So, this is the transfer function right here. It's whatever multiplies the input by, to give you the output. And let me clear my fraction. So, I get one over RC Jomega plus one. That is my transfer function right there. And if I want to find the magnitude of that transfer function. It is C one, over the magnitude of the denominator, which is the real part squared. Plus the imaginary part squared. And then the square root of that. And the angle. Well, I've got, I determined the denominator. So, it's going to be minus the imaginary part, over the real part. The R10 of that. Now, if you're not familiar with these formulas for finding magnitude and angle of complex numbers. You might look at the extra video that I made on working with complex numbers. Otherwise, if you've got a calculator, you can work with a calculator. But the thing is, this is not numbers, this is a formula. So, it's hard to plug it in a calculator, unless your calculator does symbolic manipulation. So, again, if you need some background. Go back and watch the extra video I had on working with complex numbers. Now, I want to show how to work with that transfer function, to be able to find this output. These are the formulas that we just came up with before, the two important ones, this one and this one. And let's give specific numbers here, say that's 2000 ohms. And that's ten, microferods. So we've got an R and a C. I want to keep omega, as a variable, all right. So, this means that R times C. Which we need for that formula, is equal to .02. All right, so I can calculate what H is, and what theta or what this angle is, for different frequencies. So, I'm going to make a little chart here for different frequencies. And this is one of the nice things about the transfer function is, I can compute this for different frequencies. Different amplitude inputs and outputs. So, let's say that- I have to calculate H of Omega for that particular- This particular value of RC, and this particular frequency. So, I plug these numbers in here, and I get that the magnitude is 0.83. And that the angle, when I plug in those numbers in here is- If I calculate the angle, When I plug the numbers in here, I calculate the angle to be -11.3 degrees. And, if I look at a different frequency of 500, I plug in 500 here, with that RC, I get 0.49. And, plug in into the angle formula, I get -84.2 degrees. So, this is now a formula that I can use to find out what my output amplitude is, given my input amplitude. So, my output amplitude is equal to the input amplitude, times H magnitude. So, suppose AN is equal to two. So, if an input amplitude is say, two, then it's two times the magnitude at that frequency. So, two times 0.83 is equal to 1.66 and then two times this magnitude is equal to .98. And, if I figure out what the output phase is, so that is the phase of the output. It's the input, so let's say just for simplicity, the input phase was zero. The output phase is equal to the input phase Plus H at that frequency, the angle of H. So, the output is just the input, or the transfer function phase, because the input phase was 0. So what does this mean? That means that if my input is- Two times the cosign of 10t, then my output, so this is my input. Then my output, is going to be equal to 1.66 cosign of 10t, same frequency but minus that face angle. Okay, and if my input is two times a cosign of 500t, that means my output- is .98 cos(500t-84.2 degrees). So, the bottom line is, that we took this transfer function, we found the transfer function for this circuit. And we found a formula for the magnitude and the angle. And we used that formula to find the output amplitude and the output phase for different frequencies. So, I didn't have to go back and do the impedance method, every time I changed the frequency. I found instead, a formula that relates the output amplitude to the input amplitude. The output phase to the input phase, and we showed how to apply it. So the key concepts here, are that we introduce the transfer function as a way of characterizing my circuit. So, that it showed the input to output behavior, in terms of the phasors of those inputs and outputs. We defined the transfer function, as the ratio of the output phaser to the input phaser. And breaking this around a little bit, we found that we could relate the output amplitude to the input amplitude. With that magnitude of the transfer function. And over here, we can relate the output phase to the input phase. And going back to the original circuit, when we put the original circuit input, in terms of a sinusoid. We see that, the output is also a sine wave, at that same frequency and the output amplitude is found using this formula. The output phase is found using this formula. All right, thank you very much.